Question

If $653xy$ is exactly divisible by 80, then find the value of $\left( x+y \right)$.
A. 2
B. 3
C. 4
D. 6

Answer 1

Hint: First, split 80 as $a\times b$ such that $a,b$ are coprime. Now find the number $653xy$ such that it satisfies the divisibility values of both the numbers, $a,b$. From this find the values of $x,y$ and find their sum to get the answer.

Complete step-by-step answer:
Here, we are given that the number $653xy$ is exactly divisible by 80. We have to find the value of $\left( x+y \right)$. We know that if a number, let us say N is divisible by another number $m=a\times b$, then N will be divisible by $a,b$ as well where $a,b$ are coprime. For example, 70 is divisible by 10, then 70 would be divisible by 5 and 2 both also, because $10=5\times 2$ and 5 and 2 are coprime to each other. Now let us consider our question. Here we are given a number $653xy$ where, $x,y$ are tens and units digit numbers and we have to find its sum.
We have been given that $653xy$ is divisible by 80, so $653xy$ would be divisible by 16 and 5 as well because $80=16\times 5$ and 16 and 5 are coprime. We know that according to the divisibility rule of number 5, the number divisible by 5 must end with either 0 or 5. So, we get $y$ as 0 or 5. As we know that $653xy$ is divisible by 80, so it would be divisible by 10 also, so from 0 and 5, we select the value of $y$ as 0. So, finally we get, $y=0\ldots \ldots \ldots \left( i \right)$
We know that according to the divisibility rule of numbers in the form of ${{2}^{n}}$, then the number formed by the last four digits of the given number must be divisible by ${{2}^{n}}$. For example, if $8={{2}^{3}},n=3$, is divisible by 2778512, then the number formed by the last three digits of the number (512) would be divisible by ${{2}^{3}}=8$. So, for the number $653xy$ to be divisible by $16={{2}^{4}}$, then the number formed by its last four digits, $53xy$ must be divisible by 16.
We know that $y=0$ from equation (i). So, $53xy$ or $53x0$ must be divisible by 16. Now by hit and trial, we will find the number divisible by 16.
For $x=1$, we get number 5310. This is not divisible by 16, as we get a remainder of 14.
For $x=2$, we get number 5320. This is not divisible by 16, as we get a remainder of 8.
For $x=3$, we get number 5330. This is not divisible by 16, as we get a remainder of 2.
For $x=4$, we get number 5340. This is not divisible by 16, as we get a remainder of 12.
For $x=5$, we get number 5350. This is not divisible by 16, as we get a remainder of 6.
For $x=6$, we get number 5360. This is divisible by 16, as we get a remainder of 0.
So, we get $x=6$ and $y=0$. So, we get, $x+y=6+0=6$. So, the number is 65360.
Therefore, the correct answer is option D.

Note: Looking at the question, most of the students think that it is 653 multiplied by $x$ multiplicity, but actually $x,y$ are the digits at units and tens place of the number, $653xy$. Also, the students can verify their answer by dividing the number, that is 65360 by 80 to check if it is divisible by 80 or not. And two numbers are coprime when their highest common factor is 1.